= Solution
Fix any point $p$ and use a <normal orthonormal frame> there. Such frames exist: in <geodesic normal coordinates>, $g_{ij}(p)=\delta_{ij}$, $\partial_kg_{ij}(p)=0$ and the <Christoffel symbols> vanish at $p$. Applying the <Gram-Schmidt process> smoothly to the coordinate fields gives identity coefficients at $p$ and zero first derivatives there, so it produces the required frame. Reverse one frame vector if necessary to match the <orientation>.
Because the <Levi-Civita connection> is torsion-free, $[E_j,E_k](p)=(\nabla_{E_j}E_k-\nabla_{E_k}E_j)(p)=0$. The supplied formula for the <exterior derivative> of a one-form gives
$$
d\omega_i(E_j,E_k)(p)
=E_j(\delta_{ik})(p)-E_k(\delta_{ij})(p)-\omega_i([E_j,E_k](p))=0.
$$
Thus $d\omega_i(p)=0$, and the graded <Leibniz rule> implies $d\theta_i(p)=0$. Differentiate the formula in part (b):
$$
d\nu(p)=\sum_i(-1)^{i+1}df_i(p)\wedge\theta_i(p).
$$
Since $df_i=\sum_jE_j(f_i)\omega_j$, every summand with $j\ne i$ vanishes because $\omega_j$ already occurs in $\theta_i$. The remaining term has $\omega_i\wedge\theta_i=(-1)^{i-1}\omega_g$, so both signs cancel. Hence
$$
d\nu(p)=\sum_i E_i(f_i)(p)\omega_g(p)
=(\operatorname{div}X)(p)\omega_g(p).
$$
The point was arbitrary, proving the global <exterior derivative of contracted volume form> identity
$$
\boxed{d\nu=(\operatorname{div}X)\omega_g.}
$$
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